JEE Mathematics Linear Programming MCQs Set 01

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Multiple Choice Questions

Question 1: The solution set of the constraints x + 2y ≥ 11, 3x + 4y ≤ 30, 2x + 5y ≤ 30 and x ≥ 0, y ≥ 0 includes the point
(a) (2, 3)
(b) (3, 2)
(c) (3, 4)
(d) (4, 3)
Show Answer & Explanation

Answer: (c) (3, 4)

Explanation:
1. Test each point. (2, 3): 2 + 6 = 8 < 11 fails. (3, 2): 3 + 4 = 7 < 11 fails. (4, 3): 4 + 6 = 10 < 11 fails.
2. (3, 4): 3 + 8 = 11 ≥ 11, 9 + 16 = 25 ≤ 30 and 6 + 20 = 26 ≤ 30, so all constraints hold.

Teacher's Note:
1. Substitute the points in all the inequalities.

Question 2: Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. The minimum value of F occurs at
(a) (0, 2) only
(b) (3, 0) only
(c) the mid-point of the line segment joining the points (0, 2) and (3, 0)
(d) any point on the line segment joining the points (0, 2) and (3, 0)
Show Answer & Explanation

Answer: (d) any point on the line segment joining the points (0, 2) and (3, 0)

Explanation:
1. F at the corners: (0, 2): 12, (3, 0): 12, (6, 0): 24, (6, 8): 72, (0, 5): 30.
2. The minimum 12 occurs at two adjacent corners (0, 2) and (3, 0); the objective line 4x + 6y = 12 contains the whole edge joining them.
3. So the minimum occurs at every point of that segment.

Teacher's Note:
1. Evaluate F at each corner point.

Question 3: The solution set of the following system of inequations: x + y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1, is
(a) bounded region
(b) unbounded region
(c) only one point
(d) empty set
Show Answer & Explanation

Answer: (c) only one point

Explanation:
1. From x + y ≤ 3 we get y ≤ 3 − x. Then 3x + 4y ≤ 3x + 4(3 − x) = 12 − x.
2. 3x + 4y ≥ 12 forces 12 − x ≥ 12, so x ≤ 0; with x ≥ 0 we get x = 0 and then y = 3.
3. The point (0, 3) satisfies all the conditions, so the solution set is a single point.

Teacher's Note:
1. Combine the first two inequalities.

Question 4: Every gram of wheat provides 0.1 g of proteins and 0.25 g of carbohydrates. The corresponding values of rice are 0.05 g and 0.5 g respectively. Wheat costs Rs. 4 per kg and rice Rs. 6. The minimum daily requirements of proteins and carbohydrates for an average child are 50 g and 200 g respectively. In what quantities (in grams) should wheat and rice be mixed in the daily diet to provide minimum daily requirement of proteins and carbohydrates at minimum cost?
(a) 400, 200
(b) 300, 400
(c) 200, 400
(d) 400, 300
Show Answer & Explanation

Answer: (a) 400, 200

Explanation:
1. Let x g of wheat and y g of rice. Constraints: 0.1x + 0.05y ≥ 50 (i.e. 2x + y ≥ 1000) and 0.25x + 0.5y ≥ 200 (i.e. x + 2y ≥ 800), x, y ≥ 0.
2. Corner points: (800, 0), (400, 200), (0, 1000). Cost = (4x + 6y)/1000 Rs.: 3.2, 2.8 and 6.0 respectively.
3. The minimum cost is at x = 400, y = 200.

Teacher's Note:
1. Form the constraints and test the corner points.

Question 5: The solution of the linear programming problem: maximise z = 3x1 + 5x2 subject to 3x1 + 2x2 ≤ 18, x1 ≤ 4, x2 ≤ 6, x1 ≥ 0, x2 ≥ 0 is
(a) x1 = 2, x2 = 0, z = 6
(b) x1 = 2, x2 = 6, z = 36
(c) x1 = 4, x2 = 3, z = 27
(d) x1 = 4, x2 = 6, z = 42
Show Answer & Explanation

Answer: (b) x1 = 2, x2 = 6, z = 36

Explanation:
1. The feasible corner points are (0, 0), (4, 0), (4, 3), (2, 6) and (0, 6); the point (4, 6) violates 3x1 + 2x2 ≤ 18.
2. z at these: 0, 12, 27, 36, 30.
3. The maximum is 36 at (2, 6).

Teacher's Note:
1. Check the corner points; (4, 6) is infeasible.

Question 6: The maximum value of z = 3x + 4y subject to the conditions x + y ≤ 40, x + 2y ≤ 60, x, y ≥ 0 is
(a) 130
(b) 120
(c) 40
(d) 140
Show Answer & Explanation

Answer: (d) 140

Explanation:
1. The corner points are (0, 0), (40, 0), (20, 20) and (0, 30).
2. z: 0, 120, 140, 120.
3. The maximum value is 140.

Teacher's Note:
1. Find the corner points and evaluate z.

Question 7: Consider x/2 + y/4 ≥ 1 and x/3 + y/2 ≤ 1, x, y ≥ 0. Then the number of possible solutions is
(a) Zero
(b) Unique
(c) Infinite
(d) None of these
Show Answer & Explanation

Answer: (c) Infinite

Explanation:
1. The inequalities are 2x + y ≥ 4 and 2x + 3y ≤ 6 with x, y ≥ 0.
2. The point (2, 0) satisfies both, and so do points like (1.8, 0.5): 2x + y = 4.1 ≥ 4 and 2x + 3y = 5.1 ≤ 6.
3. The feasible region is a triangle with non-zero area, so there are infinitely many solutions.

Teacher's Note:
1. Check whether the feasible region has an interior.

Question 8: A shop-keeper deals in the sale of TVs and VCPs. He has Rs. 5.2 lacs to invest and has space for only 50 pieces. A TV costs Rs. 20,000 and a VCP costs Rs. 8,000. From a TV and a VCP he earns a profit of Rs. 1500 and Rs. 800 respectively. Assuming that he sells all the items that he purchases, the maximum profit (in Rs.) he can earn is equal to
(a) 60,000
(b) 55,000
(c) 51,000
(d) 47,000
Show Answer & Explanation

Answer: (d) 47,000

Explanation:
1. Constraints: x + y ≤ 50 and 20000x + 8000y ≤ 520000, i.e. 5x + 2y ≤ 130. Profit = 1500x + 800y.
2. Corner points: (26, 0): 39000; (10, 40): 15000 + 32000 = 47000; (0, 50): 40000.
3. The maximum profit is Rs. 47,000, obtained by buying 10 TVs and 40 VCPs.

Teacher's Note:
1. Form the two constraints and compare profits at the corners.

Question 9: The constraints −x1 + x2 ≤ 1, −x1 + 3x2 ≤ 9, x1, x2 ≥ 0 define
(a) Bounded feasible space
(b) Unbounded feasible space
(c) Both bounded and unbounded feasible space
(d) None of these
Show Answer & Explanation

Answer: (b) Unbounded feasible space

Explanation:
1. Take x2 = 0 and let x1 grow: both constraints hold (−x1 ≤ 1 and −x1 ≤ 9) for every x1 ≥ 0.
2. So the feasible region extends indefinitely: it is unbounded.

Teacher's Note:
1. See whether any variable can grow without limit.

Question 10: A company manufactures two types of products A and B. The storage capacity of its godown is 100 units. Total investment amount is Rs. 30,000. The cost price of A and B are Rs. 400 and Rs. 900 respectively. Suppose all the products have sold and per unit profit is Rs. 100 and Rs. 120 through A and B respectively. If x units of A and y units of B are to be produced, then two linear constraints and the iso-profit line are respectively
(a) x + y = 100; 4x + 9y = 300, 100x + 120y = c
(b) x + y ≤ 100; 4x + 9y ≤ 300, x + 2y = c
(c) x + y ≤ 100; 4x + 9y ≤ 300, 100x + 120y = c
(d) x + y ≤ 100; 9x + 4y ≤ 300, x + 2y = c
Show Answer & Explanation

Answer: (c) x + y ≤ 100; 4x + 9y ≤ 300, 100x + 120y = c

Explanation:
1. Storage: x + y ≤ 100.
2. Investment: 400x + 900y ≤ 30000, i.e. 4x + 9y ≤ 300 (with costs Rs. 400 and Rs. 900 per unit).
3. Profit line: 100x + 120y = c.

Teacher's Note:
1. Constraints are inequalities; the iso-profit line is an equation.

Question 11: Maximise Z = 3x + 5y, subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
(a) 20 at (1, 0)
(b) 30 at (0, 6)
(c) 37 at (4, 5)
(d) 33 at (6, 3)
Show Answer & Explanation

Answer: (c) 37 at (4, 5)

Explanation:
1. Corner points: (0, 0), (7, 0), (6, 3), (4, 5), (0, 6).
2. Z: 0, 21, 33, 37, 30.
3. The maximum is 37 at (4, 5).

Teacher's Note:
1. Find all corner points of the feasible region.

Question 12: A wholesale merchant wants to start the business of cereal with Rs. 24000. Wheat is Rs. 400 per quintal and rice is Rs. 600 per quintal. He has capacity to store 200 quintal cereal. He earns the profit Rs. 25 per quintal on wheat and Rs. 40 per quintal on rice. If he stores x quintal rice and y quintal wheat, then for maximum profit, the objective function is
(a) 25x + 40y
(b) 40x + 25y
(c) 400x + 600y
(d) \( \frac{400}{40}x+\frac{600}{25}y \)
Show Answer & Explanation

Answer: (b) 40x + 25y

Explanation:
1. Rice gives Rs. 40 per quintal and he stores x quintals of rice: 40x.
2. Wheat gives Rs. 25 per quintal and he stores y quintals of wheat: 25y.
3. The profit to be maximised is Z = 40x + 25y.

Teacher's Note:
1. Match each commodity with its own profit rate.

Question 13: The maximum value of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is
(a) 10
(b) 26
(c) 35
(d) 70
Show Answer & Explanation

Answer: (c) 35

Explanation:
1. Corner points: (0, 0), (7, 0), (4, 3), (0, 5).
2. z: 0, 35, 26, 10.
3. The maximum is 35.

Teacher's Note:
1. Evaluate z at every corner point.

Question 14: For the constraints of a linear optimising function z = x1 + x2, given by x1 + x2 ≤ 1, 3x1 + x2 ≥ 3 and x1, x2 ≥ 0,
(a) There are two feasible regions
(b) There are infinite feasible regions
(c) There is no feasible region
(d) None of these.
Show Answer & Explanation

Answer: (d) None of these.

Explanation:
1. From the first constraint x2 ≤ 1 − x1, and from the second x2 ≥ 3 − 3x1, so 3 − 3x1 ≤ 1 − x1, i.e. x1 ≥ 1.
2. With x2 ≥ 0 the first constraint gives x1 ≤ 1, so x1 = 1 and x2 = 0.
3. The feasible region is the single point (1, 0), so none of the first three statements holds.

Teacher's Note:
1. Combine the two constraints to bound x1.

Question 15: The maximum value of z = 2x + 5y subject to the constraints 2x + 5y ≤ 10, x + 2y ≥ 1, x − y ≤ 4, x ≥ y ≥ 0, occurs at
(a) exactly one point
(b) exactly two points
(c) infinitely many points
(d) None of these
Show Answer & Explanation

Answer: (c) infinitely many points

Explanation:
1. The objective line 2x + 5y = z is parallel to the boundary 2x + 5y = 10, so the maximum value 10 is reached along the whole portion of that boundary which lies in the feasible region.
2. That portion is a line segment (for example it contains (2.5, 1) and (3, 0.8)), so the maximum occurs at infinitely many points.

Teacher's Note:
1. Compare the objective line with the constraint lines.

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Practice MCQs for JEE Mathematics Linear Programming

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